Solve y=-2x^2-8x+1

$$y = - 2 x ^ { 2 } - 8 x + 1$$

Solve for x

$$x=-\frac{\sqrt{18-2y}}{2}-2$$

$$x=\frac{\sqrt{18-2y}}{2}-2\text{, }y\leq 9$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$-2x^{2}-8x+1=y$$

Subtract y from both sides.

$$-2x^{2}-8x+1-y=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, -8 for b, and 1-y for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\left(-2\right)\left(1-y\right)}}{2\left(-2\right)}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\left(-2\right)\left(1-y\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-\left(-8\right)±\sqrt{64+8\left(1-y\right)}}{2\left(-2\right)}$$

Multiply 8 times 1-y.

$$x=\frac{-\left(-8\right)±\sqrt{64+8-8y}}{2\left(-2\right)}$$

Add 64 to 8-8y.

$$x=\frac{-\left(-8\right)±\sqrt{72-8y}}{2\left(-2\right)}$$

Take the square root of 72-8y.

$$x=\frac{-\left(-8\right)±2\sqrt{18-2y}}{2\left(-2\right)}$$

The opposite of -8 is 8.

$$x=\frac{8±2\sqrt{18-2y}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{8±2\sqrt{18-2y}}{-4}$$

Now solve the equation x=\frac{8±2\sqrt{18-2y}}{-4} when ± is plus. Add 8 to 2\sqrt{18-2y}.

$$x=\frac{2\sqrt{18-2y}+8}{-4}$$

Divide 8+2\sqrt{18-2y} by -4.

$$x=-\frac{\sqrt{18-2y}}{2}-2$$

Now solve the equation x=\frac{8±2\sqrt{18-2y}}{-4} when ± is minus. Subtract 2\sqrt{18-2y} from 8.

$$x=\frac{-2\sqrt{18-2y}+8}{-4}$$

Divide 8-2\sqrt{18-2y} by -4.

$$x=\frac{\sqrt{18-2y}}{2}-2$$

The equation is now solved.

$$x=-\frac{\sqrt{18-2y}}{2}-2 x=\frac{\sqrt{18-2y}}{2}-2$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$-2x^{2}-8x+1=y$$

Subtract 1 from both sides.

$$-2x^{2}-8x=y-1$$

Divide both sides by -2.

$$\frac{-2x^{2}-8x}{-2}=\frac{y-1}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{-8}{-2}x=\frac{y-1}{-2}$$

Divide -8 by -2.

$$x^{2}+4x=\frac{y-1}{-2}$$

Divide y-1 by -2.

$$x^{2}+4x=\frac{1-y}{2}$$

Divide 4, the coefficient of the x term, by 2 to get 2. Then add the square of 2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+4x+2^{2}=\frac{1-y}{2}+2^{2}$$

Square 2.

$$x^{2}+4x+4=\frac{1-y}{2}+4$$

Add \frac{-y+1}{2} to 4.

$$x^{2}+4x+4=\frac{9-y}{2}$$

Factor x^{2}+4x+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+2\right)^{2}=\frac{9-y}{2}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+2\right)^{2}}=\sqrt{\frac{9-y}{2}}$$

Simplify.

$$x+2=\frac{\sqrt{18-2y}}{2} x+2=-\frac{\sqrt{18-2y}}{2}$$

Subtract 2 from both sides of the equation.

$$x=\frac{\sqrt{18-2y}}{2}-2 x=-\frac{\sqrt{18-2y}}{2}-2$$

Solve for y

$$y=1-8x-2x^{2}$$